Example 2: eA
Compute eA, where
A = 2 -2 1
-1 3 -1
-2 -4 3
Solution:

eA = T eD T -1 ; Diagonalize A:

We first use the equation det (A - I ) = 0 to find the eigenvalues:

det (A - I ) = 2 - -2 1
-1 3 - -1
-2 -4 3 -
-Add the 2nd row to the 1st row:
= 1 - 1 - 0
-1 3 - -1
-2 -4 3 -
= ( 1 - ) 1 1 0
-1 3 - -1
-2 -4 3 -
-Take ( 1 - ) from the 1st row as a coefficient
-Add the 1st row to the 2nd row and multiplied by 2 to the 3rd row
= (1-) 1 1 0
0 4 - -1
0 -2 3 -
= (1-) 1 1 0
0 2 - 2 -
0 -2 3 -
= (1-)(2-) 1 1 0
0 1 1
0 -2 3 -
-Add the 3rd row to the 2nd row
-Take ( 2 - ) from the 2nd row as a coefficient
-Add the 2nd row multiplied by 2 to the 3rd row
= ( 1 - ) ( 2 - ) 1 1 0
0 1 1
0 0 5 -
= ( 1 - ) ( 2 - ) ( 5 - ) 1 1 0 = 0
0 1 1
0 0 1
-Take ( 5 - ) from the 3rd row as a coefficient
-The determinant of the upper triangular matrix
is = 1·1·1 = 1
=> 1 = 1, 2 = 2, 3 = 5 (eigenvalues)
=> We have
D = 1 0 0
0 2 0
0 0 5

Eigenvectors:

Using the equation (A - I ) = 0, we find the eigenvectors of A:

1 = 1: 1 = t ( 0, 1, 2 ), t C, t 0
2 = 2: 2 = t ( -1, 1, 2 ), t C, t 0
3 = 6: 3 = t ( 1, -1, 1 ), t C, r 0

Choose i's as follows:
1 = ( 0, 1, 2 ), 2 = ( -1, 1, 2 ), 3 = ( 1, -1, 1 )
and form the matrix T:

T = 0 -1 -1
1 1 1
2 2 -1
We have
T -1 = 3 3 0
-3 -2 1
0 -2 1
so we now calculate
TeDT -1 = 0 -1 -1
1 1 1
2 2 -1
e1 0 0
0 e2 0
0 0 e5
3 3 0
-3 -2 1
0 -2 1
= 3e2 2e2 + 2e5 e2 + e5
3e - 3e2 3e + 2e2 + 2e5 -e2 + e5
6e - 6e2 6e - 4e2 - 2e5 2e2 + e5
7,4 -94,0 47,0
-4,7 96,7 -47,0
-9,3 -103,4 54,4

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